Sitting arrangement – 3 men and their wives, 2 rows

combinatoricsprobability

Three men and their wives sit randomly on 6 seats in 2 rows (each row has 3 seats).

  1. What’s the number of different ways they can sit?
  2. What is the probability that all the men sit in one row?
  3. What is the probability that every man sits facing his wife?
  4. What is the probability that only one man sits facing his wife? (I believe "at least one man" would do. Else it would be too complicated?)

I attempted solving the questions. The first one is surely right, but for the next two after too many tries, I solved them programmatically, that is – recursively listing all permutations and filtering the ones that fulfill the required condition.

My solution:

  1. (6P6)=720
  2. 72/720 = 1/10
  3. 48/720 = 1/15
  4. Couldn't solve it.

Can anyone assist on how to solve them mathematically? There should be an easy way that I'm missing out, the way I solved them probably gets me the right answer but it's not logical for a human to do.

Best Answer

  1. You are correct
  2. There are two possible rows for all the men to sit on, and after choosing this row, there are $6$ ways to place the men in this row, and $6$ ways to place the women in the other row. So, we get $$\frac{2\cdot6\cdot6}{6!}=\frac1{10}$$
  3. We simply choose which column each couple should be in, and which row the man sits in, so there are $3!\cdot8$ choices. So, we get $$\frac{3!\cdot8}{6!}=\frac1{15}$$
  4. For this to happen, we first pick the couple that sits across from each other ($3$ choices), and then pick which column they do it in ($3$ choices), and which row the man is in ($2$ choices), and then allocate the rest of the seats. This is $4!-8$, since when one of the remaining couples sits across from each other, so does the other, so we count $8$ possibilities when this happens (which row is the man in for each couple, and which couple in which column). So, we get $$\frac{9\cdot16\cdot2}{6!}=\frac25$$
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